Question

A crossover trial is a type of experiment used to compare two drugs. Subjects take one drug for a period of time and then switch to the other. The responses of the subjects are then compared using matched pair methods. In an experiment to compare two pain relievers, seven subjects took one pain reliever for two weeks and then switched to the other. They rated their pain level from 1 to 10, with larger numbers representing higher levels of pain. Can you conclude that the mean response differs between the two drugs?
Subject
1 2 3 4 5 6 7
Drug A 6 3 4 5 7 1 4
Drug B 5 1 5 5 5 2 2

Answer 1

Answer:

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-0.714 -0}{\frac{1.38}{\sqrt{7}}}=-1.369$

$df=n-1=7-1=6$

$p_v =2*P(t_{(6)}$

We see that the p value is higher than the ususal significance levels commonly used of 1% or 5% so then we can conclude that we FAIL to reject the null hypothesis, and there is not enough evidence to conclude that we have a different response between the two drugs

Step-by-step explanation:

We have the following info given by the problem

Subject  1 2 3 4 5 6 7

Drug A  6 3 4 5 7 1 4

Drug B  5 1 5 5 5 2 2

x=value for drug A , y = value for drug B

x: 6 3 4 5 7 1 4  

y: 5 1 5 5 5 2 2

We want to verify if the mean response differs between the two drugs then  the system of hypothesis for this case are:

Null hypothesis: $\mu_y- \mu_x = 0$

Alternative hypothesis: $\mu_y -\mu_x \neq 0$

We can begin calculating the difference $d_i=y_i-x_i$ and we obtain this:

d: -1, -2, 1, 0, -2, 1, -2

Now we can calculate the mean difference  

$\bar d= \frac{\sum_{i=1}^n d_i}{n}=-0.714$

Now we can find the the standard deviation for the differences, and we got:

$s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}}=1.38$

And now we can calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-0.714 -0}{\frac{1.38}{\sqrt{7}}}=-1.369$

Now we can find the degrees of freedom given by:

$df=n-1=7-1=6$

We can calculate the p value, since we have a two tailed test the p value is given by:

$p_v =2*P(t_{(6)}$

We see that the p value is higher than the ususal significance levels commonly used of 1% or 5% so then we can conclude that we FAIL to reject the null hypothesis, and there is not enough evidence to conclude that we have a different response between the two drugs